Page 216 - Applied statistics and probability for engineers

194 Chapter 5/Joint Probability Distributions

Because the X’s are independent,
f x x , , x n ) = f X ( )⋅ f X ( ) ⋅ ⋅ ( x n )
…
( 1
, 2
1 x 1 2 x 2 f X n
and one may write
∞ ∞ ∞
x dx 1 ∫
M t ( ) = ∫ e f X ( ) e f X ( ) ∫ e f Xn ( )
tx n
x dx 2 ⋯
tx 1
tx 2
Y
n
x dx n
2
1
−∞ 1 −∞ ∞ 2 −∞
t ( ) t ( ⋅ ) ⋯ ⋅ M Xn ( t)
= M X 1 ⋅ M X 2
For the case when the X’s are discrete, we would use the same approach replacing integrals
with summations.
Equation 5-35 is particularly useful. In many situations we need to ind the distribution of

the sum of two or more independent random variables, and often this result makes the problem
very easy. This is illustrated in the following example.
Example 5-38 Distribution of a Sum of Poisson Random Variables Suppose that X 1 and X 2 are two inde-
pendent Poisson random variables with parameters λ 1 and λ 2 , respectively. Determine the prob-
ability distribution of Y = X + X 2 .
1
The moment-generating function of a Poisson random variable with parameter λ is
t
M t) = e λ e ( −1 )
(
X
t
t
t ( ) = e λ 1 ( e 1)− t ( ) = e λ 2 ( e 1)−
so the moment-generating functions of X 1 and X 2 are M X 1 and M X 2 , respectively. Using
Equation 5-35, the moment-generating function of Y = X + X 2 is
1
t
t
t
M t) = M X ( ) 2 t) e 1 λ e ( −1 ) λ 2 e ( −1 ) = e ( 1 λ + 2 )( e −1 )
t M X ( =
λ
e
(
Y
1
which is recognized as the moment-generating function of a Poisson random variable with parameter λ 1 + λ 2 . There-
fore, the sum of two independent Poisson random variables with parameters λ 1 and λ 2 is a Poisson random variable
with parameter equal to the sum λ 1 + λ 2 .
EXERCISES FOR SECTION 5-6
Problem available in WileyPLUS at instructor’s discretion.
Tutoring problem available in WileyPLUS at instructor’s discretion
5-91. A random variable X has the discrete uniform distribution (b) Use M t ( ) to i nd the mean and variance of the Poisson
X
…
2
(
f x) = 1 , x = 1 , , , m random variable.
5-93. The geometric random variable X has probability
m
distribution
(a) Show that the moment-generating function is − x−1
2
f x ( ) (= 1 p) p, x = 1 , ,…
e ( − e )
t
tm
1
(
M t) = Show that the moment-generating function is
m( − e ) t) pe t
X
t
1
M X ( = − t
(b) Use M t ( ) to i nd the mean and variance of X. 1 − 1 ( p e )
X
5-92. A random variable X has the Poisson distribution Use M X ( t) to ind the mean and variance of X.
−λ
e λ x 5-94. The chi-squared random variable with k degrees of free-
(
, ,…
f x) = , x = 0 1 dom has moment-generating function M X t( ) (= −1 2 t)− k / 2.
x!
Suppose that X 1 and X 2 are independent chi-squared random
(a) Show that the moment-generating function is
variables with k 1 and k 2 degrees of freedom, respectively. What
t
(
M t) = e λ e ( −1 ) is the distribution of Y = X + X 2 ?
1
X